Let S be a set. S* is the power set of S. Prove there can't be any onto mapping from S->S*

My approach:
1. Let f be a onto mapping from S->S*
2. Define a set X as follows:
X={s| s belong to S AND s doesn't belong to f(s)}
3. Claim X is not in the image of 'f'
Let X=f(s) for some s in S.
It is easy to show element 's' will belong exactly to ONE of X or f(s)
Hence X =/= f(s)

The proposition is true for any set, and this proof is valid for any set, whether it be uncountable, infinite or whatever else you can think of. The crucial fact is that

is a valid subset of for any set A good question to ask youself is "why wouldn't this be true for an uncountabe set?" You should eventually be convinced that there is no reason.

N.b. It is the stipulation that in the definition of which rescues us from paradoxes such as Russel's Paradox.

Nov 11th 2009, 08:48 AM

aman_cc

Quote:

Originally Posted by nimon

The proposition is true for any set, and this proof is valid for any set, whether it be uncountable, infinite or whatever else you can think of. The crucial fact is that

is a valid subset of for any set A good question to ask youself is "why wouldn't this be true for an uncountabe set?" You should eventually be convinced that there is no reason.

N.b. It is the stipulation that in the definition of which rescues us from paradoxes such as Russel's Paradox.

Thanks Nimon. Yes - I too find my proof pretty convincing. But I'm still a novice at all this - so wanted to be sure.

Initially I constructed X by defining it in terms of s1,s2,,,,sn,.... where each si was in S. This, I felt, somehow limits the proof to countable sets. So, I modified the definition of X.

I want to be sure I understand your comments well.

"A good question to ask youself is "why wouldn't this be true for an uncountabe set?" You should eventually be convinced that there is no reason."

Is there more to it than what I stated in pt 3 of my post ? -
3. Claim X is not in the image of 'f'
Let X=f(s) for some s in S.
It is easy to show element 's' will belong exactly to ONE of X or f(s)
Hence X =/= f(s)

It is not to difficult to prove that the mapping given by is a bijection. Therefore . Therefore you must show that and cannot be equipotent. Can you see a slightly easier way to proceed now?

Nov 12th 2009, 12:16 AM

aman_cc

Quote:

Originally Posted by Drexel28

There is a slightly easier way.

It is not to difficult to prove that the mapping given by is a bijection. Therefore . Therefore you must show that and cannot be equipotent. Can you see a slightly easier way to proceed now?

Thanks Drexel28m for your post. Please forgive my ignorance, I have not understood a few things here.

1. - What does this mean, I mean I don't understand the definitions here? Please feel free to point me where I can read it up.

2.

This is also not clear to me

3. What is equipotent?

Nov 12th 2009, 06:43 AM

Drexel28

Quote:

Originally Posted by aman_cc

Thanks Drexel28m for your post. Please forgive my ignorance, I have not understood a few things here.

1. - What does this mean, I mean I don't understand the definitions here? Please feel free to point me where I can read it up.

2.

This is also not clear to me

3. What is equipotent?

Don't feel ignorant. Just a difference in notation.

Definitions:

- power set of

. In other words it is the set of all functions that map from into.

- This is just the charceristic function of . It equals 1 if and if not.

equipotent- Two sets are equipotent (denoted ) if there exists a bijection between the two, i.e. .

Therefore what one means when they say " given by " what they are really saying is "define a mapping between the power set of and the set of all functions that map into by a subset of (lets say E) is mapped to the charceristic function of that subset ( ). Doing this we can easily prove that . Therefore instead of proving that and aren't equipotent you merely need to show that and aren't equipotent (this is because equipotence is an equivalance relation). This is actually an easier task.

Here is how. I will give you the outline and you fill in the "why?" in the last step.

Let be some injective mapping from to . Then , and since we see that . Now define a second mapping by . Then clearly but (why?). Therefore we may conclude that there is no surjection between and . Consequently, there is no surjection between and . Therefore

Remark: Realizing that given by is an injection, we may conclude that . Combining this with teh above we may definitively say that

I hope that helps.

Nov 12th 2009, 06:49 AM

aman_cc

Quote:

Originally Posted by Drexel28

Don't feel ignorant. Just a difference in notation.

Definitions:

- power set of

. In other words it is the set of all functions that map from into.

- This is just the charceristic function of . It equals 1 if and if not.

equipotent- Two sets are equipotent (denoted ) if there exists a bijection between the two, i.e. .

Therefore what one means when they say " given by " what they are really saying is "define a mapping between the power set of and the set of all functions that map into by a subset of (lets say E) is mapped to the charceristic function of that subset ( ). Doing this we can easily prove that . Therefore instead of proving that and aren't equipotent you merely need to show that and aren't equipotent (this is because equipotence is an equivalance relation). This is actually an easier task.

Here is how. I will give you the outline and you fill in the "why?" in the last step.

Let be some injective mapping from to . Then , and since we see that . Now define a second mapping by . Then clearly but (why?). Therefore we may conclude that there is no surjection between and . Consequently, there is no surjection between and . Therefore

Remark: Realizing that given by is an injection, we may conclude that . Combining this with teh above we may definitively say that

I hope that helps.

Really appreciate your help and effort. I will read your response carefully and get back for any further questions. Thanks

Nov 13th 2009, 03:23 AM

aman_cc

Quote:

Originally Posted by Drexel28

Don't feel ignorant. Just a difference in notation.

Definitions:

- power set of

. In other words it is the set of all functions that map from into.

- This is just the charceristic function of . It equals 1 if and if not.

equipotent- Two sets are equipotent (denoted ) if there exists a bijection between the two, i.e. .

Therefore what one means when they say " given by " what they are really saying is "define a mapping between the power set of and the set of all functions that map into by a subset of (lets say E) is mapped to the charceristic function of that subset ( ). Doing this we can easily prove that . Therefore instead of proving that and aren't equipotent you merely need to show that and aren't equipotent (this is because equipotence is an equivalance relation). This is actually an easier task.

Here is how. I will give you the outline and you fill in the "why?" in the last step.

Let be some injective mapping from to . Then , and since we see that . Now define a second mapping by . Then clearly but (why?). Therefore we may conclude that there is no surjection between and . Consequently, there is no surjection between and . Therefore

Remark: Realizing that given by is an injection, we may conclude that . Combining this with teh above we may definitively say that

I hope that helps.

Let me try to answer your 'why' part. Most of your proof is clear to me. However, I'm trouble understanding the way is defined. I think this is due to the way you have used and in the definition.
So I'm gonna define it the way I have understood and you can tell me if I got it correct.

I would also assume is a bijection and thus a surjection from . You I guess assumed that is an injection. I have a dis-connect here as well.

Now (as defined by me) , definitely

As is a surjection so such that .

But consider and . Obviously if one is 1 the other is 0. Hence . Thus can't be a surjection and thus can't be a bijection.

Am I correct?

Nov 17th 2009, 08:09 PM

Drexel28

Quote:

Originally Posted by aman_cc

Let me try to answer your 'why' part. Most of your proof is clear to me. However, I'm trouble understanding the way is defined. I think this is due to the way you have used and in the definition.
So I'm gonna define it the way I have understood and you can tell me if I got it correct.

I would also assume is a bijection and thus a surjection from . You I guess assumed that is an injection. I have a dis-connect here as well.

Now (as defined by me) , definitely

As is a surjection so such that .

But consider and . Obviously if one is 1 the other is 0. Hence . Thus can't be a surjection and thus can't be a bijection.